Symmetry assignment of spin-flip states in EOM

A fundamental question for the EOM experts…

The Q-Chem manual notes the following:

It is a symmetry of a transition rather than that of a target state which is specified in excited state calculations. The symmetry of the target state is a product of the symmetry of the reference state and the transition. For closed-shell molecules, the former is fully symmetric and the symmetry of the target state is the same as that of transition, however, for open-shell references this is not so.

Does this same consideration apply to spin-flip EOM calculations? Or can the symmetry of the SF states be directly read off of the output?

Thanks!
Ryan